Unexpected result from ArcLengthDetermining which rule NIntegrate selects automaticallyFinding minimum fly-by radius between Mars and spacecraft from interpolating functionCoarse-graining in numerical integrationsNIntegrate fails to converge around a value out of integration rangeA 1D numerical integral Mathematica cannot compute, from physicsDifferential Equation with Numerically Integrated Boundary ConditionsDifferents results of the intersection area between two regions when using the function “Area” and the function “NIntegrate”Issue with boundary Integration of FEM numerical solution (interpolation function)How to get the most accurate volume of a special solid?Numerical solution of 3 dim integral with singularity
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Unexpected result from ArcLength
Determining which rule NIntegrate selects automaticallyFinding minimum fly-by radius between Mars and spacecraft from interpolating functionCoarse-graining in numerical integrationsNIntegrate fails to converge around a value out of integration rangeA 1D numerical integral Mathematica cannot compute, from physicsDifferential Equation with Numerically Integrated Boundary ConditionsDifferents results of the intersection area between two regions when using the function “Area” and the function “NIntegrate”Issue with boundary Integration of FEM numerical solution (interpolation function)How to get the most accurate volume of a special solid?Numerical solution of 3 dim integral with singularity
$begingroup$
I want to determine the arc lenght of a parametric curve $C: x(t),y(t) = cos(t)^p , sin(t)^p $ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.
I set up the following function of $p$:
L[p_] := ArcLength[Cos[t]^p, Sin[t]^p, t, 0, Pi/2,
Method -> "NIntegrate", MaxRecursion -> 20]
For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708
.
For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100]
results in 1.30603
. Not close to 2
(it's not even bigger than Pi/2
).
Plotting, results in the following:
Plot[L[p], p, 0, 1]
Any ideas? I'm running 11.0.0.0
numerical-integration
$endgroup$
add a comment |
$begingroup$
I want to determine the arc lenght of a parametric curve $C: x(t),y(t) = cos(t)^p , sin(t)^p $ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.
I set up the following function of $p$:
L[p_] := ArcLength[Cos[t]^p, Sin[t]^p, t, 0, Pi/2,
Method -> "NIntegrate", MaxRecursion -> 20]
For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708
.
For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100]
results in 1.30603
. Not close to 2
(it's not even bigger than Pi/2
).
Plotting, results in the following:
Plot[L[p], p, 0, 1]
Any ideas? I'm running 11.0.0.0
numerical-integration
$endgroup$
1
$begingroup$
I get a warning fromNIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluateL
for smallp
.
$endgroup$
– MarcoB
1 hour ago
$begingroup$
@MarcoB I don't get any warnings when evaluatingL[1/100]
$endgroup$
– Ivan
1 hour ago
add a comment |
$begingroup$
I want to determine the arc lenght of a parametric curve $C: x(t),y(t) = cos(t)^p , sin(t)^p $ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.
I set up the following function of $p$:
L[p_] := ArcLength[Cos[t]^p, Sin[t]^p, t, 0, Pi/2,
Method -> "NIntegrate", MaxRecursion -> 20]
For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708
.
For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100]
results in 1.30603
. Not close to 2
(it's not even bigger than Pi/2
).
Plotting, results in the following:
Plot[L[p], p, 0, 1]
Any ideas? I'm running 11.0.0.0
numerical-integration
$endgroup$
I want to determine the arc lenght of a parametric curve $C: x(t),y(t) = cos(t)^p , sin(t)^p $ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.
I set up the following function of $p$:
L[p_] := ArcLength[Cos[t]^p, Sin[t]^p, t, 0, Pi/2,
Method -> "NIntegrate", MaxRecursion -> 20]
For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708
.
For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100]
results in 1.30603
. Not close to 2
(it's not even bigger than Pi/2
).
Plotting, results in the following:
Plot[L[p], p, 0, 1]
Any ideas? I'm running 11.0.0.0
numerical-integration
numerical-integration
edited 1 hour ago
Henrik Schumacher
56.6k577156
56.6k577156
asked 1 hour ago
IvanIvan
1,612821
1,612821
1
$begingroup$
I get a warning fromNIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluateL
for smallp
.
$endgroup$
– MarcoB
1 hour ago
$begingroup$
@MarcoB I don't get any warnings when evaluatingL[1/100]
$endgroup$
– Ivan
1 hour ago
add a comment |
1
$begingroup$
I get a warning fromNIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluateL
for smallp
.
$endgroup$
– MarcoB
1 hour ago
$begingroup$
@MarcoB I don't get any warnings when evaluatingL[1/100]
$endgroup$
– Ivan
1 hour ago
1
1
$begingroup$
I get a warning from
NIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluate L
for small p
.$endgroup$
– MarcoB
1 hour ago
$begingroup$
I get a warning from
NIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluate L
for small p
.$endgroup$
– MarcoB
1 hour ago
$begingroup$
@MarcoB I don't get any warnings when evaluating
L[1/100]
$endgroup$
– Ivan
1 hour ago
$begingroup$
@MarcoB I don't get any warnings when evaluating
L[1/100]
$endgroup$
– Ivan
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
$endgroup$
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
1 hour ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
1 hour ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
1 hour ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
1 hour ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength[]
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
1 hour ago
|
show 9 more comments
$begingroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
]
Plot[L[p], p, 0.001, 1]
$endgroup$
add a comment |
$begingroup$
Seems to be a precision thing.
L[p_] = Cos[t]^p, Sin[t]^p
ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]
1.99447959240474567...
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
$endgroup$
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
1 hour ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
1 hour ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
1 hour ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
1 hour ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength[]
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
1 hour ago
|
show 9 more comments
$begingroup$
Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
$endgroup$
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
1 hour ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
1 hour ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
1 hour ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
1 hour ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength[]
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
1 hour ago
|
show 9 more comments
$begingroup$
Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
$endgroup$
Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
edited 1 hour ago
answered 1 hour ago
mjwmjw
5779
5779
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
1 hour ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
1 hour ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
1 hour ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
1 hour ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength[]
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
1 hour ago
|
show 9 more comments
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
1 hour ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
1 hour ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
1 hour ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
1 hour ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength[]
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
1 hour ago
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
1 hour ago
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
1 hour ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
1 hour ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
1 hour ago
2
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
1 hour ago
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
1 hour ago
1
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
1 hour ago
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
1 hour ago
1
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is the
ArcLength[]
returning wrong results? Seems that in both cases Mathematica is undersampling ...$endgroup$
– mjw
1 hour ago
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is the
ArcLength[]
returning wrong results? Seems that in both cases Mathematica is undersampling ...$endgroup$
– mjw
1 hour ago
|
show 9 more comments
$begingroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
]
Plot[L[p], p, 0.001, 1]
$endgroup$
add a comment |
$begingroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
]
Plot[L[p], p, 0.001, 1]
$endgroup$
add a comment |
$begingroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
]
Plot[L[p], p, 0.001, 1]
$endgroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
]
Plot[L[p], p, 0.001, 1]
edited 1 hour ago
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
56.6k577156
56.6k577156
add a comment |
add a comment |
$begingroup$
Seems to be a precision thing.
L[p_] = Cos[t]^p, Sin[t]^p
ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]
1.99447959240474567...
$endgroup$
add a comment |
$begingroup$
Seems to be a precision thing.
L[p_] = Cos[t]^p, Sin[t]^p
ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]
1.99447959240474567...
$endgroup$
add a comment |
$begingroup$
Seems to be a precision thing.
L[p_] = Cos[t]^p, Sin[t]^p
ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]
1.99447959240474567...
$endgroup$
Seems to be a precision thing.
L[p_] = Cos[t]^p, Sin[t]^p
ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]
1.99447959240474567...
answered 39 mins ago
Bill WattsBill Watts
3,4811620
3,4811620
add a comment |
add a comment |
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1
$begingroup$
I get a warning from
NIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluateL
for smallp
.$endgroup$
– MarcoB
1 hour ago
$begingroup$
@MarcoB I don't get any warnings when evaluating
L[1/100]
$endgroup$
– Ivan
1 hour ago